xslt sorting from two different X paths

sen1953 · September 3rd, 2013, 01:01 PM

This is works fine its working but...

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0"exclude-result-prefixes="xs saxon"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:saxon="http://saxon.sf.net/">
<xsl:template match="/">
<xsl:for-each select="root/E/F/A | root/d/A">
<xsl:sort select="."data-type="number"order="ascending"/>
<xsl:value-of select="."/>
</xsl:for-each>
</xsl:template>

</xsl:stylesheet>

But in this code, instead of <xsl:sort select="A" > I given <xsl:sort select="." , Its not accepting as <xsl:sort select="A" >

1.Why its behaving like this?
My requirement is, I got to combine two different logic from two xpaths .
2.so I cant use one <xsl:value-of select="."/> for the two xpaths like "root/E/F/A | root/d/A">
I got to transform like this
<xsl:value-of select=first path of 'A'>
<xsl:value-of select=second path of 'A'>
I cant have one value of select for these two xpaths.

{

logic for path root/E/F/A
1 john london
3 leon london
}

{
logic for path2 root/d/A
6 neo chicaco
8 raj london
9 seon london
}

In this case I cant use one value-of select these blocks

Cant I use two value of select for two xpath ?

mhkay · September 3rd, 2013, 01:15 PM

You say this code is working, and it looks fine to me. Your explanation as to why you can't use this code is incoherent. If it works, why isn't it OK?

Remember:

* the xsl:for-each should select the set of nodes to be sorted

* the xsl:sort select expression should evaluate the sort key, using each item to be sorted as the context item. So if the item being sorted is an A element, then xsl:sort select="A" won't work, unless your A elements have children called A, which doesn't appear to be the case.

sen1953 · September 3rd, 2013, 03:32 PM

My problem is my requiremt is like below

Code:

 
root> 
 
 
<E> 
<F>

        <A>3</A>
</F>
</E>
<E>
<F>
       <A>8</A>
</F>
</E>
<E>
<F>
       <A>2</A>
</F>
</E>
<d>
<b>
       <A>5</A>
</b>
<b>
      <A>1</A>
</b>
<b>
      <A>9</A>
 </b>
<b>
     <A>7</A>
</b>
</d>
 
 
 
</root>

I got to sort based on 'A' elements. But not all ,particularly for these paths root/E/F/A and root/d/b/A
E nodes have 1 to many and b also 1 to many..

for these I cant apply these xsl as below

Code:

 
<?xml version="1.0" encoding="UTF-8"?>

<xsl:stylesheet version="1.0"exclude-result-prefixes="xs saxon"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:saxon="http://saxon.sf.net/">
<xsl:template match="/">
<xsl:for-each select="root/E/F/A | root/d/A">
<xsl:sort select="."data-type="number"order="ascending"/>
<xsl:value-of select="."/>
</xsl:for-each>
</xsl:template>
 
 
</xsl:stylesheet>

how to sort out this issue ?
<xsl:for-each select="root/E/F/A">
<xsl:sort select="."data-type="number"order="ascending"/>
<xsl:value-of select="."/>
</xsl:for-each>
<xsl:for-each select=" root/d/A">
<xsl:sort select="."data-type="number"order="ascending"/>
<xsl:value-of select="."/>
</xsl:for-each>

Can we combine two for-each then do sorting Is it possible or not.Thanks in advance

mhkay · September 3rd, 2013, 04:04 PM

You keep changing the problem!

What output do you want from this latest input?

sen1953 · September 3rd, 2013, 04:24 PM

My required output is only
1235789

If you give solution by combine two for-each and then sorting , will it be great for my requirement. I completed my all work without sorting .That is my problem. Let us assume, each for-each will have complex logic not only
number. for example my requirement will be like this

Code:

2 john london             this will from different path
5 Dillon london
7 london 6sep             this will from different path l
9 madrid 4aug

so i need flexibility on sorting. Thanks in advance...

mhkay · September 3rd, 2013, 04:56 PM

>My required output is only 1235789

Well you can achieve that using

Code:

<xsl:for-each select="//A">
  <xsl:sort select="."/>
  <xsl:value-of select="."/>
</xsl:for-each>

But you haven't explained why your own "working" code isn't good enough. You aren't explaining your requirements very clearly.

sen1953 · September 3rd, 2013, 05:38 PM

Yes you are correct. I didnt explain my problem clearly..

this is my input xml

Code:

<?xml version="1.0" encoding="UTF-8"?>
<accountList>
<previousAccount>
<account>
<lastName>NASH</lastName>
<accountStatus>REMOVED</accountStatus>
<accNo>08</accNo>
</account>
<account>
<lastName>ADOGA</lastName>
<accountStatus>REMOVED</accountStatus>
<accNo>01</accNo>
</account>
<account>
<lastName>LUCAS</lastName>
<accountStatus>HOLD</accountStatus>
<accNo>09</accNo>
</account>
<account>
<lastName>DONALD</lastName>
<accountStatus>HOLD</accountStatus>
<accNo>21</accNo>
</account>
<account>
<accountStatus>HOLD</accountStatus>
<lastName>LONDON</lastName>
<accNo>24</accNo>
</account>
</previousAccount>
<account>
<Title>Mr</Title>
<firstName>RICHARD</firstName>
<lastName>JOHN</lastName>
<city>london</city>
<accNo>02</accNo>
</account>
<account>
<Title>Mr</Title>
<firstName>xxx</firstName>
<lastName>JOHN</lastName>
<city>London</city>
<accNo>17</accNo>
</account>
<account>
<Title>Mr</Title>
<firstName>HEWIT</firstName>
<lastName>JOHN</lastName>
<city>LONDON</city>
<accNo>20</accNo>
</account>
<account>
<Title>Mr</Title>
<firstName>xxx</firstName>
<lastName>JOHN</lastName>
<city>LONDON</city>
<accNo>21</accNo>
</account>
<account>
<Title>Mr</Title>
<firstName>KEVIN</firstName>
<lastName>PETE</lastName>
<city>LONDON</city>
<accNo>07</accNo>
</account>
</accountList>

my xsl code

Code:

 
 
<xsl:stylesheet version="2.0"
 
xsi:schemaLocation="http://www.w3.org/1999/XSL/Transform" 
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"

xmlns:fn="http://www.w3.org/2005/xpath-functions"
xmlns:xs="http://www.w3.org/2001/XMLSchema"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<xsl:output method="text"/>
<xsl:variable name="newline">
<xsl:text>
</xsl:text>
</xsl:variable>
 

<xsl:template match="/">
 

<xsl:for-each select="accountList/account">
<xsl:sort select="accNo"order="ascending"/><xsl:value-of select="accNo"/>
 
 

<xsl:text> </xsl:text>
<xsl:value-of select="lastName"/>
<xsl:value-of select="$newline"/>
</xsl:for-each>
 

<xsl:value-of select="$newline"/>
 

<xsl:for-each select="accountList/previousAccount/account">
<xsl:sort select="accNo"order="ascending"/>
<xsl:value-of select="accNo"/>
<xsl:text> </xsl:text>
<xsl:value-of select="accountStatus"/>
<xsl:text> </xsl:text>
<xsl:value-of select="lastName"/>
<xsl:value-of select="$newline"/>
</xsl:for-each>
 

</xsl:template>
 
 
 
 
</xsl:stylesheet>

current output without soring

Code:

 
02                        JOHN

 
07                         PETE

 
17                         JOHN

20                        JOHN
21                        JOHN
 
 

 
 
 
 
01 REMOVED    ADOGA

 
08 REMOVED     NASH

09 HOLD             LUCAS
21 HOLD            DONALD
24 HOLD            LONDON

Expected As usual I got to sortout by accno

Code:

01 REMOVED ADOGA

 
02                        JOHN

07                        PETE
08  REMOVED  NASH
09  HOLD           LUCAS
17                      JOHN
20                       JOHN
21                      JOHN
24HOLD           LONDON

Code:

That is my problem. Now you can understand my requirement as well. I tried many approach.I tried lot.. Please provide soltuion.

Thanks in advance

samjudson · September 4th, 2013, 12:13 PM

And back to my original answer (from nearly a month ago!):

Code:

<xsl:for-each select="accountList/account | accountList/previousAccount/account">
        <xsl:sort select="accNo" order="ascending"/>
        <xsl:value-of select="accNo"/>
        <!-- do stuff with the other elements -->
</xsl:for-each>

What is wrong with this? Why does this not meet your requirements?

Is it because it still requires you to do some actual work (by filling in the bit in the middle?) or for some other reason I cannot fathom?

sen1953 · September 10th, 2013, 12:17 PM

Thanks .I solved this by below approach

Code:

<xsl:for-each select="accountList/account | accountList/previousAccount/account">
<xsl:sort select="accNo" order="ascending"/>
<xsl:value-of select="accNo"/>
<!-- do stuff with the other elements -->
</xsl:for-each>

Problem is ,I did wrong xpaths. That is why its not working well.we got to concentrate more on xpaths when we go for two different xpath. I understood the concepts and I came to know how to use xpath efficiently as well.

And also we can solve this same issue by the below approach

Code:

<xsl:variable name="sortingAccNumber">
- <xsl:for-each select="accountList/account">
<xsl:sort select="accNo" /> 
<xsl:copy-of select="." /> 
</xsl:for-each>
- <xsl:for-each select="accountList/previousAccount/account">
<xsl:sort select="accNo" /> 
<xsl:copy-of select="." /> 
</xsl:for-each>
</xsl:variable>
- <xsl:for-each select="ext:node-set($sortingAccNumber)/account">
<xsl:sort select="accNo" /> 
<xsl:value-of select="accNo" /> 
</xsl:for-each>

Thanks a lot Mckay and Samjudson for your extended support